A drawer contains red and black balls. When two balls are drawn at random, the probability that they both are red is `(1)/(2)`. The number of balls in the drawer can be

Options

A. 21
B. 11
C. 4
D. 3

Answer

To solve the problem, we need to determine the total number of balls in the drawer based on the given probability that when two balls are drawn at random, both are red, which is \( \frac{1}{2} \).

### Step-by-Step Solution:

1. **Define Variables**:
Let \( p \) be the total number of balls in the drawer, and \( m \) be the number of red balls.

2. **Set Up the Probability Equation**:
The probability of drawing 2 red balls from \( m \) red balls out of \( p \) total balls is given by:
\[ P(\text{both red}) = \frac{\binom{m}{2}}{\binom{p}{2}} = \frac{1}{2} \]
where \( \binom{n}{k} \) is the binomial coefficient representing the number of ways to choose \( k \) items from \( n \) items.

3. **Express the Binomial Coefficients**:
The binomial coefficients can be expressed as:
\[ \binom{m}{2} = \frac{m(m-1)}{2} \]
\[ \binom{p}{2} = \frac{p(p-1)}{2} \]
Substituting these into the probability equation gives:
\[ \frac{\frac{m(m-1)}{2}}{\frac{p(p-1)}{2}} = \frac{1}{2} \]

4. **Simplify the Equation**:
The \( \frac{1}{2} \) cancels out, leading to:
\[ \frac{m(m-1)}{p(p-1)} = \frac{1}{2} \]
Cross-multiplying gives:
\[ 2m(m-1) = p(p-1) \]

5. **Rearranging the Equation**:
Rearranging the equation results in:
\[ 2m^2 - 2m - p^2 + p = 0 \]
This is a quadratic equation in \( m \).

6. **Using the Quadratic Formula**:
The quadratic formula \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) can be applied here:
- \( a = 2 \)
- \( b = -2 \)
- \( c = -p^2 + p \)

Thus, we have:
\[ m = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 2 \cdot (-p^2 + p)}}{2 \cdot 2} \]
Simplifying gives:
\[ m = \frac{2 \pm \sqrt{4 + 8p^2 - 8p}}{4} \]
\[ m = \frac{2 \pm \sqrt{8p^2 - 8p + 4}}{4} \]
\[ m = \frac{1 \pm \sqrt{2p^2 - 2p + 1}}{2} \]

7. **Condition for \( m \) to be an Integer**:
For \( m \) to be an integer, \( \sqrt{2p^2 - 2p + 1} \) must be an odd perfect square.

8. **Testing Possible Values for \( p \)**:
We test the options given (21, 11, 4, 3) to check if \( 2p^2 - 2p + 1 \) is a perfect square.

- **For \( p = 21 \)**:
\[ 2(21^2) - 2(21) + 1 = 882 - 42 + 1 = 841 \quad (\text{which is } 29^2) \]
- **For \( p = 11 \)**:
\[ 2(11^2) - 2(11) + 1 = 242 - 22 + 1 = 221 \quad (\text{not a perfect square}) \]
- **For \( p = 4 \)**:
\[ 2(4^2) - 2(4) + 1 = 32 - 8 + 1 = 25 \quad (\text{which is } 5^2) \]
- **For \( p = 3 \)**:
\[ 2(3^2) - 2(3) + 1 = 18 - 6 + 1 = 13 \quad (\text{not a perfect square}) \]

9. **Conclusion**:
The possible values for \( p \) that satisfy the condition are \( 21 \) and \( 4 \).

### Final Answer:
The number of balls in the drawer can be **21 or 4**.